Factoring using GCF

Question

So this Section is "solving quadratic equations by factoring", I got the hang of it, but stumped on a few. So one problem says 8y^2 -28y-60 and the answer is 4(2y+3)(y-5). I do not get how they got that though. Because for other problems like 5x^2+5x-10 I came up with 5(1x^2+1x-2) and it seems legit (Did it according to how a site taught iT)

Also troubled on 3x^2+54x+243. The GCF is 3, so I do 3(1x^2+18x+81). Either thats it or I break it down to 3(1x+2x)(1x+9).

Thanks.

Answer 1

8y² - 28y - 60, factor out common factor of "4"

4 ( 2y² - 7y - 15 ), factor using FOIL

4 ( 2y + 3 )( y - 5 )

 

Following that patter,

 

5x² - 5x - 10

5 ( x² - x - 2 )

5 ( x -2 )( x + 1)

 

You are correct to "break it down" for the third one.

Answer 2

8y² -28y -60

 

we can start by looking for any factor that is common in all terms

 

The number 4 is a factor of each term so we can factor it out

 

4(2y² - 7y -15)

 

OK now we have to find the factors of the first and last terms of the binomial and figure out which combination will give us the middle term

 

factors of 2y²

 

are 2y and y

 

factors of -15

 

-1 and 15

1 and -15

 

-3 and 5

3 and -5

 

now we can try combinations of these factors  to get -7 in the middle

 

I will leave it to you to try different combinations but the correct combination is 3 and -5

becasue 2y times -5 gives -10y and y times 3 gives 3y and the combination gives -7y

 

so our factors are

4(2y+3)(y-5)

 

hope this helps